One must always DEFINE the positive sense of all signed quantities and stick
with those definitions in writing equation models and in interpreting
results of calculations. Just as this applies to coordinates (eg: x,y,z) in
kinematics, it applies no less in this capacitor discharge situation:
1) One must specify the positive direction of the current i. You may choose
either direction as positive;
2) One must specify which capacitor plate carries the signed quantity q.
You may choose either plate for this purpose; the other plate then carries
the charge (-q).
If you are faithful to your specifications, and interpret your results
accordingly, you may then traverse the circuit in either direction and your
"loop theorem" statement will be valid.
Note if one chooses positive i in 1) to be the direction INTO the capacitor
plate chosen to carry q in 2), then it follows that i = dq/dt. If instead
one chooses positive i in 1) to be the direction OUT OF the capacitor plate
chosen to carry q in 2), then it follows that i = - dq/dt.
Note here the unspoken convention that the positive direction of i refers to
the flow of POSITIVE charge carriers.
Bob Sciamanda
Physics, Edinboro Univ of PA (Em) http://www.winbeam.com/~trebor/
trebor@winbeam.com
----- Original Message -----
From: "Michael Edmiston" <edmiston@bluffton.edu>
To: "PHYS-L Maillist" <phys-l@carnot.physics.buffalo.edu>
Sent: Saturday, February 18, 2006 5:49 PM
Subject: [Phys-l] RC Disharge Analysis
|I searched the archives to see if this has been discussed before because
| it seems it should have been. I didn't see it. My apologies if it has
| already been discussed.
|
| When analizing a circuit containing a charged capacitor, a resistor, and
| an initially-open switch, the analysis in many textbooks goes like
| this...
|
| (1) close the switch
|
| (2) apply the loop theorem and get q/C + IR = 0
|
| In some books this is written -q/C -IR = 0. Normally I would say this
| is because they went around the loop in the other direction, but as I
| explain below, it seems neither is correct.
|
| (3) Replace I with dq/dt
|
| (4) Rearrange to get dq/q = -dt/RC
|
| (5) Integrate to get q(t) = Qexp(-t/RC) where Q is the initial charge.
|
| Of course (5) is the correct answer, but (2) is not what I (and
| students) get from the loop theorem. Proper application of the loop
| theorem yields opposite signs for the two terms rather than the same
| sign. Then, following through steps 3 to 5 yields q(t) = Qexp(+t/RC)
| which is the incorrect answer.
|
| A couple sources I have do the loop theorem in step (2) as q/C - IR = 0
| or -q/C + IR = 0 depending on whether you do the loop with the current
| or opposite the current. Then they get the signs both the same by
| applying some handwaving in step (3). The handwaving goes like this...
| I = dq/dt, but since q is decreasing we have to write I = -dq/dt.
|
| It seems to me that the textbooks that get the same sign for the two
| terms from the application of the loop theorem in step (2) are just
| plain wrong. No matter which way you navigate the loop, one delta-V
| will be positive and the other negative. Indeed, when analyzing the
| charging of the circuit (with battery in the loop) most books do the
| loop theorem properly... E - q/C - IV = 0 going in the current
| direction. Serway and Jewitt actually say (for charging), "For the
| capacitor, notice that we are traveling in the direction from the
| positive plate to the negative plate; this represents a decrease in
| potential." (page 874) But then on page 876 where they analyze the
| discharge they use the same equation and just drop the E because the
| battery has been removed. The problem is, when the battery is removed
| (and replaced with a wire) the current flows the opposite direction. In
| this case (current direction) we still have a potential drop across the
| resistor, but we now have a potential increase across the capacitor (if
| we apply the same reasoning as stated in the quoted portion from page
| 874). Serway and Jewitt just seem to ignore this, i.e. they make no
| comment about which way they went around the loop and how -q/C -IR could
| possibly be zero if q,C,I,R are all unsigned numbers.
|
| One the other hand, even though I prefer saying I = -dq/dt in step (3),
| the problem with setting I = -dq/dt is that it certainly seems to
| students that some handwaving is being done just to get the correct
| answer. Indeed, when current is typically defined as dq/dt we usually
| explain that it doesn't really mean charge is changing, but rather a
| certain amount of charge dq passes a point in time dt. From that
| definition of I = dq/dt it seems difficult to justify setting it to
| minus dq/dt in the capacitor problem.
|
| Are any of you bothered by this? Do any of you have a clean way out of
| this?
|
| Michael D. Edmiston, Ph.D.
| Professor of Physics and Chemistry
| Bluffton University
| Bluffton, OH 45817
| (419)-358-3270
| edmiston@bluffton.edu
|
|
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